Chapter 1: Light — Reflection and Refraction

Solution

The first law of reflection states that the angle of incidence (∠i) is always equal to the angle of reflection (∠r). The second law states that the incident ray, reflected ray, and normal all lie in the same plane. Snell's law and the law of refraction deal with refraction, not reflection.

Solution

\(f = \frac{R}{2} = \frac{46}{2} = 23\) cm

Solution

\(f = \frac{R}{2} = \frac{55}{2} = 27\) cm

Solution

\(f = \frac{R}{2} = \frac{7}{2} = 3\) cm

Solution

\(f = \frac{R}{2} = \frac{22}{2} = 11\) cm

Solution

\(f = \frac{R}{2} = \frac{59}{2} = 29\) cm

Solution

\(f = \frac{R}{2} = \frac{35}{2} = 17\) cm

Solution

\(f = \frac{R}{2} = \frac{50}{2} = 25\) cm

Solution

\(f = \frac{R}{2} = \frac{17}{2} = 8\) cm

Solution

\(f = \frac{R}{2} = \frac{28}{2} = 14\) cm

Solution

\(f = \frac{R}{2} = \frac{8}{2} = 4\) cm

Solution

\(f = \frac{R}{2} = \frac{38}{2} = 19\) cm

Solution

\(f = \frac{R}{2} = \frac{6}{2} = 3\) cm

Solution

\(f = \frac{R}{2} = \frac{48}{2} = 24\) cm

Solution

\(f = \frac{R}{2} = \frac{19}{2} = 9\) cm

Solution

\(f = \frac{R}{2} = \frac{34}{2} = 17\) cm

Solution

\(f = \frac{R}{2} = \frac{43}{2} = 21\) cm

Solution

\(f = \frac{R}{2} = \frac{54}{2} = 27\) cm

Solution

\(f = \frac{R}{2} = \frac{57}{2} = 28\) cm

Solution

\(f = \frac{R}{2} = \frac{24}{2} = 12\) cm

Solution

\(f = \frac{R}{2} = \frac{10}{2} = 5\) cm

Solution

\(f = \frac{R}{2} = \frac{53}{2} = 26\) cm

Solution

\(f = \frac{R}{2} = \frac{21}{2} = 10\) cm

Solution

\(f = \frac{R}{2} = \frac{9}{2} = 4\) cm

Solution

\(f = \frac{R}{2} = \frac{15}{2} = 7\) cm

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{40} = 2.5\) D

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{30} = 3.33\) D

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{20} = 5.0\) D

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{25} = 4.0\) D

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{15} = 6.67\) D

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{10} = 10.0\) D

Solution

Power \(= \frac{1}{f(\text{in m})} = \frac{100}{50} = 2.0\) D

Solution

\(f = \frac{R}{2} = \frac{60}{2} = 30\) cm

Solution

\(f = \frac{R}{2} = \frac{29}{2} = 14\) cm

Solution

\(f = \frac{R}{2} = \frac{14}{2} = 7\) cm

Solution

\(f = \frac{R}{2} = \frac{20}{2} = 10\) cm

Solution

\(f = \frac{R}{2} = \frac{58}{2} = 29\) cm

Solution

\(f = \frac{R}{2} = \frac{52}{2} = 26\) cm

Solution

\(f = \frac{R}{2} = \frac{11}{2} = 5\) cm

Solution

\(f = \frac{R}{2} = \frac{23}{2} = 11\) cm

Solution

\(f = \frac{R}{2} = \frac{31}{2} = 15\) cm

Solution

\(f = \frac{R}{2} = \frac{56}{2} = 28\) cm

Solution

\(f = \frac{R}{2} = \frac{12}{2} = 6\) cm

Solution

\(f = \frac{R}{2} = \frac{40}{2} = 20\) cm

Solution

\(f = \frac{R}{2} = \frac{47}{2} = 23\) cm

Solution

\(f = \frac{R}{2} = \frac{30}{2} = 15\) cm

Solution

\(f = \frac{R}{2} = \frac{27}{2} = 13\) cm

Solution

\(f = \frac{R}{2} = \frac{16}{2} = 8\) cm

Solution

\(f = \frac{R}{2} = \frac{51}{2} = 25\) cm

Solution

\(f = \frac{R}{2} = \frac{25}{2} = 12\) cm

Solution

A concave mirror is called a converging mirror because when parallel rays of light fall on it, they converge (come together) at the focus after reflection. A convex mirror is the diverging mirror — it spreads rays apart.

Solution

For a spherical mirror, the relationship between focal length (f) and radius of curvature (R) is:

f = R / 2

f = 30 / 2 = 15 cm

Solution

Convex mirrors are used as rear-view mirrors because they always form erect and diminished images. More importantly, they have a wider field of view compared to plane mirrors, allowing the driver to see a larger area of the road behind. The images are always virtual, erect, and smaller than the object.

Solution

Glass is optically denser than air. When light enters a denser medium from a rarer medium, it bends toward the normal. Since the refractive index of glass (~1.5) is greater than air (~1.0), light slows down in glass. The speed of light in glass is approximately c/1.5 = 2 × 10⁸ m/s.

Solution

Given: f = -50 cm = -0.5 m (negative because concave lens)

Power: P = 1/f = 1/(-0.5) = -2 D

The negative sign indicates it is a diverging (concave) lens.

Solution

Regular Reflection: Occurs on smooth, polished surfaces. All reflected rays are parallel to each other, forming a clear image. Example: plane mirror, still water surface.

Diffuse Reflection: Occurs on rough, unpolished surfaces. Reflected rays scatter in different directions because the surface has micro-irregularities. Example: paper, wall, fabric.

Does diffuse reflection violate the laws of reflection?

No. Even in diffuse reflection, each individual ray obeys both laws of reflection perfectly. At each tiny point on the rough surface, the angle of incidence equals the angle of reflection. The scattering occurs because the normals at different points on the surface point in different directions due to the surface roughness.

Solution

When we look at a pencil partly immersed in water, the part below the water surface appears displaced or bent. Here's why:

Light rays from the submerged part of the pencil travel from water (denser medium) to air (rarer medium). At the water-air interface, these rays bend away from the normal (refraction from denser to rarer medium).

When these refracted rays reach our eyes, our brain traces them back in straight lines. The apparent position of the submerged part of the pencil is where these extended straight lines seem to originate — which is higher than its actual position.

Since the part above water is at its true position and the part below appears shifted upward, the pencil appears to be bent or broken at the surface. This is an optical illusion caused by refraction.

Solution

When an object is placed at the focus (F₁) of a convex lens, the refracted rays emerge parallel to the principal axis. Parallel rays never meet (or meet at infinity). Therefore, the image is formed at infinity. It is real, inverted, and highly enlarged. This is the principle behind searchlights and collimators.

Solution

A concave lens is a diverging lens. It always diverges the rays passing through it. The diverging rays, when traced back, appear to meet on the same side as the object. This produces an image that is always virtual, erect, and diminished — regardless of where the object is placed.

Solution

P = 1/f (where f is in metres)

f = 1/P = 1/2 = 0.5 m = 50 cm

Since P is positive, it is a convex (converging) lens.

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-50} = -0.8\)

Solution

\(m = -\frac{v}{u} = -\frac{-15}{-30} = -0.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-35} = -0.71\)

Solution

\(m = -\frac{v}{u} = -\frac{40}{-24} = 1.67\)

Solution

\(m = -\frac{v}{u} = -\frac{-20}{-40} = -0.5\)

Solution

\(m = -\frac{v}{u} = -\frac{20}{-45} = 0.44\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-40} = -0.75\)

Solution

\(m = -\frac{v}{u} = -\frac{20}{-24} = 0.83\)

Solution

\(m = -\frac{v}{u} = -\frac{40}{-25} = 1.6\)

Solution

\(m = -\frac{v}{u} = -\frac{40}{-30} = 1.33\)

Solution

\(m = -\frac{v}{u} = -\frac{-15}{-50} = -0.3\)

Solution

\(m = -\frac{v}{u} = -\frac{-15}{-40} = -0.38\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-60} = -0.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-60}{-50} = -1.2\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-45} = -0.67\)

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-24} = -1.67\)

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-25} = -1.6\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-24} = -1.25\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-40} = -0.62\)

Solution

\(m = -\frac{v}{u} = -\frac{30}{-50} = 0.6\)

Solution

\(m = -\frac{v}{u} = -\frac{20}{-50} = 0.4\)

Solution

\(m = -\frac{v}{u} = -\frac{-20}{-60} = -0.33\)

Solution

\(m = -\frac{v}{u} = -\frac{-60}{-30} = -2.0\)

Solution

\(m = -\frac{v}{u} = -\frac{40}{-60} = 0.67\)

Solution

\(m = -\frac{v}{u} = -\frac{60}{-24} = 2.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-50} = -0.6\)

Solution

\(m = -\frac{v}{u} = -\frac{-20}{-24} = -0.83\)

Solution

\(m = -\frac{v}{u} = -\frac{30}{-60} = 0.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-60}{-24} = -2.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-20} = -1.25\)

Solution

\(m = -\frac{v}{u} = -\frac{60}{-50} = 1.2\)

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-40} = -1.0\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-45} = -0.56\)

Solution

\(m = -\frac{v}{u} = -\frac{-20}{-15} = -1.33\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-15} = -2.0\)

Solution

\(m = -\frac{v}{u} = -\frac{60}{-30} = 2.0\)

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-45} = -0.89\)

Solution

\(m = -\frac{v}{u} = -\frac{30}{-24} = 1.25\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-35} = -0.86\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-24} = -1.04\)

Solution

\(m = -\frac{v}{u} = -\frac{40}{-45} = 0.89\)

Solution

\(m = -\frac{v}{u} = -\frac{20}{-60} = 0.33\)

Solution

\(m = -\frac{v}{u} = -\frac{-60}{-25} = -2.4\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-60} = -0.42\)

Solution

\(m = -\frac{v}{u} = -\frac{-20}{-35} = -0.57\)

Solution

\(m = -\frac{v}{u} = -\frac{-60}{-35} = -1.71\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-20} = -1.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-60} = -0.67\)

Solution

\(m = -\frac{v}{u} = -\frac{30}{-20} = 1.5\)

Solution

\(m = -\frac{v}{u} = -\frac{-25}{-30} = -0.83\)

Solution

\(m = -\frac{v}{u} = -\frac{60}{-20} = 3.0\)

Solution

\(m = -\frac{v}{u} = -\frac{-60}{-45} = -1.33\)

Solution

\(m = -\frac{v}{u} = -\frac{60}{-60} = 1.0\)

Solution

\(m = -\frac{v}{u} = -\frac{40}{-15} = 2.67\)

Solution

\(m = -\frac{v}{u} = -\frac{30}{-35} = 0.86\)

Solution

\(m = -\frac{v}{u} = -\frac{-20}{-50} = -0.4\)

Solution

\(m = -\frac{v}{u} = -\frac{60}{-15} = 4.0\)

Solution

\(m = -\frac{v}{u} = -\frac{-30}{-30} = -1.0\)

Solution

\(m = -\frac{v}{u} = -\frac{-40}{-30} = -1.33\)

Solution

When an object is placed between F and P of a concave mirror, the reflected rays diverge. If we extend them behind the mirror, they appear to meet, forming a virtual, erect, and enlarged image. This is the principle used in shaving mirrors and makeup mirrors.

Solution

Given: u = -30 cm (object is always on the left), f = -15 cm (concave mirror has negative f)

Using mirror formula:

1/v + 1/u = 1/f

1/v + 1/(-30) = 1/(-15)

1/v = -1/15 + 1/30

1/v = (-2 + 1)/30 = -1/30

v = -30 cm

The negative sign means the image is formed in front of the mirror — it is real.

Magnification:

m = -v/u = -(-30)/(-30) = -1

|m| = 1, so the image is the same size as the object. The negative sign indicates the image is inverted.

Nature: Real, inverted, same size, formed at the centre of curvature (since v = -30 cm = 2f = R).

Solution

Given: u = -20 cm, f = +10 cm (convex mirror has positive f)

Using mirror formula:

1/v + 1/u = 1/f

1/v + 1/(-20) = 1/10

1/v = 1/10 + 1/20 = (2 + 1)/20 = 3/20

v = +20/3 ≈ +6.67 cm

Positive v means the image is behind the mirror — it is virtual.

Magnification:

m = -v/u = -(20/3)/(-20) = +1/3 ≈ +0.33

Positive m means the image is erect. |m| < 1 means it is diminished (one-third the size).

Solution

Given: n = 1.5, c = 3 × 10⁸ m/s

Formula: n = c/v

Therefore: v = c/n = (3 × 10⁸) / 1.5

v = 2 × 10⁸ m/s

Light travels at two-thirds of its vacuum speed when passing through glass.

Solution

The refractive index n = c/v. A refractive index of 2.42 means:

2.42 = (speed in vacuum)/(speed in diamond)

So light travels 2.42 times slower in diamond compared to vacuum. This is why diamond bends light so dramatically and sparkles brilliantly — its high refractive index causes significant bending and internal reflections.

Solution

Given: u = -30 cm (object on left), f = +20 cm (convex lens)

Using lens formula:

1/v - 1/u = 1/f

1/v - 1/(-30) = 1/20

1/v + 1/30 = 1/20

1/v = 1/20 - 1/30 = (3 - 2)/60 = 1/60

v = +60 cm

Positive v means the image is on the right side of the lens — it is real.

Magnification: m = v/u = 60/(-30) = -2

Negative m means the image is inverted. |m| = 2 means it is twice the size of the object (enlarged).

Solution

Given: P = +1.5 D

Focal length: f = 1/P = 1/1.5 = +0.667 m ≈ +66.7 cm

Since the power is positive, this is a converging (convex) lens.

A positive power lens is prescribed for hypermetropia (long-sightedness), where the person has difficulty seeing nearby objects.

Solution

Given: P₁ = +3.5 D, P₂ = -2.5 D

Net power: P = P₁ + P₂ = 3.5 + (-2.5) = +1 D

Focal length: f = 1/P = 1/1 = +1 m = 100 cm

The combination behaves like a converging lens (positive power) of focal length 100 cm.

Solution

Ray Diagram Description:

Place the object (arrow) between C and F on the principal axis of a concave mirror.

• Ray 1: Draw a ray parallel to the principal axis from the top of the object to the mirror. After reflection, this ray passes through F.
• Ray 2: Draw a ray from the top of the object passing through F to the mirror. After reflection, this ray becomes parallel to the principal axis.

The two reflected rays meet at a point beyond C, on the same side as the object. Draw the image (inverted arrow) at this point.

Characteristics of the image:

• Position: Beyond the centre of curvature C
• Size: Enlarged (larger than the object)
• Nature: Real (formed by actual intersection of reflected rays)
• Orientation: Inverted (upside down)

Solution

A concave mirror is used in headlights because of a special property: when a light source is placed at the focus (F) of a concave mirror, the reflected rays emerge as a powerful parallel beam of light.

This is the reverse of what happens when parallel rays hit a concave mirror (they converge at F). By placing the bulb at F, we get the reverse — a strong, directed beam that illuminates the road ahead over long distances without scattering.

The bulb should be placed at the focus of the concave mirror. If the bulb is placed slightly above or below the focus, the beam spreads slightly, which is how low-beam/high-beam adjustments work in some headlight designs.

Solution

Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media and for a given colour of light.

n₁ sin i = n₂ sin r

Or: sin i / sin r = n₂/n₁

Calculation:

Refractive index of glass with respect to water:

n(glass/water) = n(glass)/n(water) = 1.52/1.33 = 1.14

This means light travels 1.14 times slower in glass than in water. When light passes from water to glass, it bends toward the normal since glass is optically denser than water.

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-25}\)

\(v = -16.67\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-30} - \frac{1}{-50}\)

\(v = -75.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-45}\)

\(v = -20.32\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-30} - \frac{1}{-30}\)

\(v = infinity\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-12} - \frac{1}{-25}\)

\(v = -23.08\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-12} - \frac{1}{-40}\)

\(v = -17.14\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-12} - \frac{1}{-35}\)

\(v = -18.26\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-35}\)

\(v = -14.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-60}\)

\(v = -12.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-16} - \frac{1}{-15}\)

\(v = 240.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-25}\)

\(v = -37.5\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-20} - \frac{1}{-50}\)

\(v = -33.33\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-25} - \frac{1}{-50}\)

\(v = -50.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-35}\)

\(v = -23.33\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-40}\)

\(v = -24.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-20}\)

\(v = -46.67\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-50}\)

\(v = -12.5\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-16} - \frac{1}{-40}\)

\(v = -26.67\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-25} - \frac{1}{-24}\)

\(v = 600.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-20} - \frac{1}{-15}\)

\(v = 60.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-30} - \frac{1}{-60}\)

\(v = -60.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-25}\)

\(v = -31.82\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-20} - \frac{1}{-30}\)

\(v = -60.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-25} - \frac{1}{-20}\)

\(v = 100.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-60}\)

\(v = -18.26\) cm

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(55^\circ) = 1.52 \times \sin r\)

\(\sin r = \frac{\sin 55^\circ}{1.52}\)

\(r = 32.6^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.44 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.44}\)

\(r = 26.5^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(50^\circ) = 1.33 \times \sin r\)

\(\sin r = \frac{\sin 50^\circ}{1.33}\)

\(r = 35.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(45^\circ) = 1.5 \times \sin r\)

\(\sin r = \frac{\sin 45^\circ}{1.5}\)

\(r = 28.1^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(55^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 55^\circ}{1.55}\)

\(r = 31.9^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(30^\circ) = 1.6 \times \sin r\)

\(\sin r = \frac{\sin 30^\circ}{1.6}\)

\(r = 18.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(25^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 25^\circ}{1.55}\)

\(r = 15.8^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(60^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 60^\circ}{1.45}\)

\(r = 36.7^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(60^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 60^\circ}{1.55}\)

\(r = 34.0^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(55^\circ) = 1.33 \times \sin r\)

\(\sin r = \frac{\sin 55^\circ}{1.33}\)

\(r = 38.0^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(30^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 30^\circ}{1.55}\)

\(r = 18.8^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(60^\circ) = 1.33 \times \sin r\)

\(\sin r = \frac{\sin 60^\circ}{1.33}\)

\(r = 40.6^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(30^\circ) = 1.7 \times \sin r\)

\(\sin r = \frac{\sin 30^\circ}{1.7}\)

\(r = 17.1^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(50^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 50^\circ}{1.45}\)

\(r = 31.9^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(25^\circ) = 1.52 \times \sin r\)

\(\sin r = \frac{\sin 25^\circ}{1.52}\)

\(r = 16.1^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(25^\circ) = 1.5 \times \sin r\)

\(\sin r = \frac{\sin 25^\circ}{1.5}\)

\(r = 16.4^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(25^\circ) = 1.33 \times \sin r\)

\(\sin r = \frac{\sin 25^\circ}{1.33}\)

\(r = 18.5^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(50^\circ) = 1.5 \times \sin r\)

\(\sin r = \frac{\sin 50^\circ}{1.5}\)

\(r = 30.7^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(30^\circ) = 1.5 \times \sin r\)

\(\sin r = \frac{\sin 30^\circ}{1.5}\)

\(r = 19.5^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(25^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 25^\circ}{1.45}\)

\(r = 16.9^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.45}\)

\(r = 26.3^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.55}\)

\(r = 24.5^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(60^\circ) = 1.7 \times \sin r\)

\(\sin r = \frac{\sin 60^\circ}{1.7}\)

\(r = 30.6^\circ\)

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -45 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{45}\)

v = 13.85 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -25 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{25}\)

v = 12.5 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -60 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{60}\)

v = 17.65 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -25 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{25}\)

v = 11.11 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -40 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{40}\)

v = 8.0 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -40 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{40}\)

v = 13.33 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -35 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{35}\)

v = 12.73 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -35 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{35}\)

v = 14.58 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -15 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{15}\)

v = 8.57 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -35 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{35}\)

v = 8.94 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -15 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{15}\)

v = 6.0 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -15 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{15}\)

v = 6.67 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -45 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{45}\)

v = 16.07 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -35 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{35}\)

v = 7.78 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -50 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{50}\)

v = 16.67 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -30 cm (sign convention), f = 15 cm

\(\frac{1}{v} = \frac{1}{15} + \frac{1}{30}\)

v = 10.0 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -45 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{45}\)

v = 9.47 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -20 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{20}\)

v = 7.5 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -35 cm (sign convention), f = 15 cm

\(\frac{1}{v} = \frac{1}{15} + \frac{1}{35}\)

v = 10.5 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -50 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{50}\)

v = 8.33 cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{-15}\)

\(f = -30.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{15}\)

\(f = 6.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{25}\)

\(f = 9.38\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{30}\)

\(f = 10.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{20}\)

\(f = 10.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{-20}\)

\(f = 20.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{-50}\)

\(f = 21.43\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{30}\)

\(f = 15.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{15}\)

\(f = 8.57\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{-15}\)

\(f = -60.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{-15}\)

\(f = -37.5\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{-20}\)

\(f = -100.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{-50}\)

\(f = 12.5\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{10}\)

\(f = 6.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{-30}\)

\(f = 60.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{-40}\)

\(f = 40.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{30}\)

\(f = 12.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{25}\)

\(f = 7.14\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{30}\)

\(f = 13.64\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{30}\)

\(f = 7.5\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{-40}\)

\(f = 24.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{20}\)

\(f = 12.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{20}\)

\(f = 11.11\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-8} - \frac{1}{-40}\)

\(v = -10.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-20}\)

\(v = -20.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-60}\)

\(v = -20.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-45}\)

\(v = -30.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-20} - \frac{1}{-25}\)

\(v = -100.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-60}\)

\(v = -25.71\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-30}\)

\(v = -45.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-45}\)

\(v = -22.5\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-25} - \frac{1}{-60}\)

\(v = -42.86\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-30}\)

\(v = -30.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-30}\)

\(v = -15.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-16} - \frac{1}{-25}\)

\(v = -44.44\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-35}\)

\(v = -37.06\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-30} - \frac{1}{-35}\)

\(v = -210.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-30}\)

\(v = -26.25\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-8} - \frac{1}{-60}\)

\(v = -9.23\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-50}\)

\(v = -28.13\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-20} - \frac{1}{-35}\)

\(v = -46.67\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-35}\)

\(v = -26.25\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-40}\)

\(v = -32.73\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-16} - \frac{1}{-45}\)

\(v = -24.83\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-15} - \frac{1}{-50}\)

\(v = -21.43\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-10} - \frac{1}{-15}\)

\(v = -30.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-18} - \frac{1}{-24}\)

\(v = -72.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-20} - \frac{1}{-60}\)

\(v = -30.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-12} - \frac{1}{-15}\)

\(v = -60.0\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-16} - \frac{1}{-35}\)

\(v = -29.47\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-8} - \frac{1}{-35}\)

\(v = -10.37\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-14} - \frac{1}{-40}\)

\(v = -21.54\) cm

Solution

Mirror formula: \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)

\(\frac{1}{v} = \frac{1}{-16} - \frac{1}{-30}\)

\(v = -34.29\) cm

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(35^\circ) = 1.6 \times \sin r\)

\(\sin r = \frac{\sin 35^\circ}{1.6}\)

\(r = 21.0^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(50^\circ) = 1.52 \times \sin r\)

\(\sin r = \frac{\sin 50^\circ}{1.52}\)

\(r = 30.3^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(55^\circ) = 1.7 \times \sin r\)

\(\sin r = \frac{\sin 55^\circ}{1.7}\)

\(r = 28.8^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(35^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 35^\circ}{1.45}\)

\(r = 23.3^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(55^\circ) = 1.5 \times \sin r\)

\(\sin r = \frac{\sin 55^\circ}{1.5}\)

\(r = 33.1^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(55^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 55^\circ}{1.45}\)

\(r = 34.4^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(45^\circ) = 1.44 \times \sin r\)

\(\sin r = \frac{\sin 45^\circ}{1.44}\)

\(r = 29.4^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(25^\circ) = 1.7 \times \sin r\)

\(\sin r = \frac{\sin 25^\circ}{1.7}\)

\(r = 14.4^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(35^\circ) = 1.7 \times \sin r\)

\(\sin r = \frac{\sin 35^\circ}{1.7}\)

\(r = 19.7^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(45^\circ) = 1.6 \times \sin r\)

\(\sin r = \frac{\sin 45^\circ}{1.6}\)

\(r = 26.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.33 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.33}\)

\(r = 28.9^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(35^\circ) = 1.52 \times \sin r\)

\(\sin r = \frac{\sin 35^\circ}{1.52}\)

\(r = 22.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.7 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.7}\)

\(r = 22.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(45^\circ) = 1.45 \times \sin r\)

\(\sin r = \frac{\sin 45^\circ}{1.45}\)

\(r = 29.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(60^\circ) = 1.44 \times \sin r\)

\(\sin r = \frac{\sin 60^\circ}{1.44}\)

\(r = 37.0^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.52 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.52}\)

\(r = 25.0^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(35^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 35^\circ}{1.55}\)

\(r = 21.7^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(45^\circ) = 1.55 \times \sin r\)

\(\sin r = \frac{\sin 45^\circ}{1.55}\)

\(r = 27.1^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(50^\circ) = 1.6 \times \sin r\)

\(\sin r = \frac{\sin 50^\circ}{1.6}\)

\(r = 28.6^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(30^\circ) = 1.52 \times \sin r\)

\(\sin r = \frac{\sin 30^\circ}{1.52}\)

\(r = 19.2^\circ\)

Solution

Snell's Law: \(n_1 \sin i = n_2 \sin r\)

\(1 \times \sin(40^\circ) = 1.5 \times \sin r\)

\(\sin r = \frac{\sin 40^\circ}{1.5}\)

\(r = 25.4^\circ\)

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -50 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{50}\)

v = 9.68 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -25 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{25}\)

v = 7.14 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -60 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{60}\)

v = 15.0 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -20 cm (sign convention), f = 20 cm

\(\frac{1}{v} = \frac{1}{20} + \frac{1}{20}\)

v = 10.0 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -40 cm (sign convention), f = 15 cm

\(\frac{1}{v} = \frac{1}{15} + \frac{1}{40}\)

v = 10.91 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -60 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{60}\)

v = 8.57 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -60 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{60}\)

v = 10.0 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -20 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{20}\)

v = 11.11 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -30 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{30}\)

v = 8.57 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -45 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{45}\)

v = 8.18 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -25 cm (sign convention), f = 12 cm

\(\frac{1}{v} = \frac{1}{12} + \frac{1}{25}\)

v = 8.11 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -30 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{30}\)

v = 7.5 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -50 cm (sign convention), f = 15 cm

\(\frac{1}{v} = \frac{1}{15} + \frac{1}{50}\)

v = 11.54 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -15 cm (sign convention), f = 25 cm

\(\frac{1}{v} = \frac{1}{25} + \frac{1}{15}\)

v = 9.38 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -20 cm (sign convention), f = 15 cm

\(\frac{1}{v} = \frac{1}{15} + \frac{1}{20}\)

v = 8.57 cm

Solution

Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

u = -20 cm (sign convention), f = 10 cm

\(\frac{1}{v} = \frac{1}{10} + \frac{1}{20}\)

v = 6.67 cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{-15}\)

\(f = 30.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{25}\)

\(f = 13.64\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{-20}\)

\(f = 60.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{-50}\)

\(f = 33.33\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{15} + \frac{1}{-30}\)

\(f = 30.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{25}\)

\(f = 12.5\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{-30}\)

\(f = 15.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{-40}\)

\(f = 66.67\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{-20}\)

\(f = -60.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{25} + \frac{1}{10}\)

\(f = 7.14\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{15}\)

\(f = 10.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{-30}\)

\(f = infinity\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{25}\)

\(f = 11.11\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{20}\)

\(f = 6.67\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{30} + \frac{1}{-40}\)

\(f = 120.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{10} + \frac{1}{10}\)

\(f = 5.0\) cm

Solution

\(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}\)

\(= \frac{1}{20} + \frac{1}{-20}\)

\(f = infinity\) cm

Solution

Given: h = 5 cm, u = -20 cm, f = -15 cm

Step 1: Find v using mirror formula

1/v + 1/u = 1/f

1/v + 1/(-20) = 1/(-15)

1/v = -1/15 + 1/20

1/v = (-4 + 3)/60 = -1/60

v = -60 cm

Negative v means the image is formed 60 cm in front of the mirror — it is real.

Step 2: Find magnification and image height

m = -v/u = -(-60)/(-20) = -3

h' = m × h = -3 × 5 = -15 cm

The negative sign of h' means the image is inverted (below the principal axis).

Answer: The image is formed 60 cm in front of the mirror, is 15 cm tall, real, inverted, and 3 times enlarged.

Solution

Given: h = 4 cm, u = -16 cm, f = +24 cm

Note: The object is placed between F and O (u < f), so we expect a virtual image.

Step 1: Lens formula

1/v - 1/u = 1/f

1/v - 1/(-16) = 1/24

1/v + 1/16 = 1/24

1/v = 1/24 - 1/16 = (2 - 3)/48 = -1/48

v = -48 cm

Negative v means the image is on the same side as the object — it is virtual.

Step 2: Magnification and image size

m = v/u = (-48)/(-16) = +3

h' = m × h = 3 × 4 = +12 cm

Positive m and positive h' mean the image is erect. The image is 12 cm tall — 3 times enlarged.

Nature: Virtual, erect, enlarged, on the same side as the object.

Solution

Given: u = -10 cm, m = -3 (real image is inverted, so m is negative for 3× magnification)

Step 1: Find v using magnification

m = -v/u

-3 = -v/(-10)

-3 = v/10

v = -30 cm

Negative v confirms the image is real (in front of the mirror).

Step 2: Find f using mirror formula

1/v + 1/u = 1/f

1/(-30) + 1/(-10) = 1/f

-1/30 - 1/10 = 1/f

(-1 - 3)/30 = 1/f

-4/30 = 1/f

f = -30/4 = -7.5 cm

The focal length is 7.5 cm. The negative sign confirms it is a concave mirror.

Solution

Given: h = 5 cm, u = -10 cm, R = +30 cm (convex mirror)

f = R/2 = +30/2 = +15 cm

Step 1: Mirror formula

1/v + 1/u = 1/f

1/v + 1/(-10) = 1/15

1/v = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6

v = +6 cm

Positive v means image is behind the mirror — virtual.

Step 2: Magnification and image height

m = -v/u = -(6)/(-10) = +0.6

h' = m × h = 0.6 × 5 = +3 cm

Positive m and h' confirm the image is erect. |m| < 1 so it is diminished.

Answer: Virtual, erect image formed 6 cm behind the mirror, 3 cm tall.

Solution

Given: u = -15 cm, f = -10 cm (concave lens has negative f)

Lens formula:

1/v - 1/u = 1/f

1/v - 1/(-15) = 1/(-10)

1/v + 1/15 = -1/10

1/v = -1/10 - 1/15 = (-3 - 2)/30 = -5/30 = -1/6

v = -6 cm

Negative v for a lens means the image is on the same side as the object — it is virtual.

Magnification:

m = v/u = (-6)/(-15) = +0.4

Positive m means the image is erect. |m| = 0.4 < 1, so the image is diminished (40% of original size).

Solution

Given: f₁ = +25 cm = +0.25 m (convex), f₂ = -10 cm = -0.10 m (concave)

Power of individual lenses:

P₁ = 1/f₁ = 1/0.25 = +4 D

P₂ = 1/f₂ = 1/(-0.10) = -10 D

Net power:

P = P₁ + P₂ = 4 + (-10) = -6 D

Focal length of combination:

f = 1/P = 1/(-6) = -0.1667 m = -16.67 cm

Since the net power is negative, the combination acts as a diverging (concave) lens. The concave lens dominates because its power magnitude (10 D) exceeds that of the convex lens (4 D).